1
$\begingroup$

TL;DR: I'm attempting to make a black hole rendering, and I like the current size of the event-horizon but I want to reduce the curvature further away from it. But it seems like $g_{uv}$ only depends on $r_s$ given an object's current coordinates. Are there any other parameters I can change?


Currently I have a spherical coordinate system $(ct, r, \theta, \phi)$ that extends 10 Unity meters in radius from the origin, with an $r_s$ around 0.5. When a ray intersects the spherical coordinate system, I express it's direction vector in terms of $e_r, e_\theta, e_\phi$, and calculate a component for $e_t$ by requiring $\frac{d}{d\lambda}\cdot\frac{d}{d\lambda} = 0$. Then from it's initial spherical coords/direction, I run 15,000 iterations of Euler's method on the geodesic equations, and convert back to Unity cartesian coordinates, dropping $\frac{d(ct)}{d\lambda}$ to send a new raycast. This is my current result:

enter image description here

I don't want to change the size of the event-horizon $r_s$, but I've am trying to reduce the curvature far away from the event horizon so that the region in pink smoothly connects to the surrounding flat space. As well, I am trying to get the white-light region outlined in orange closer to the event horizon, so it looks more like what I'm used to seeing.


Both $g_{\theta\theta}$ and $g_{\phi\phi}$ are only dependent on the current coordinates.

$g_{tt} = 1-\frac{r_s}{r}$ and $g_{rr} = -g_{tt}^{-1}$, so it seems like the metric tensor is only a function of the Schwarzschild radius. So if I want to keep $r_s$ constant, there does not seem to be a parameter to reduce distance curvature.

I've tried putting random coefficients in the geodesic equations, but adjustments seem to have the effect of reducing distance curvature.

I've largely ignored $ct$, just including it for the sake of the geodesic equations and dropping it later. But I don't think changing it would have an affect, because $r_s$ is a function of $c^2$ already, which I am trying to hold constant. Nothing else in the code is in terms of G or M because $r_s$ is known in advanced.

Given that Newton's gravity is proportional to $\frac{1}{r^2}$, it would almost seem like you can scale the current geometry, but not change how it evolves over large distances.


Are there any parameters that I'm not considering? Or is this not doable?

Thanks!


Javier's answer worked:

enter image description here

$\endgroup$

1 Answer 1

1
$\begingroup$

The Schwarzschild metric only depends on $r_s$, there are no further parameters that you could tweak. I can think of two options:

  • This is the most obvious one: you're seeing a sharp transition because you established a sharp transition between the curved and flat regions. You could just expand the curved region to cover the whole viewing area (which is more realistic anyway).

  • Alternatively, you could modify the metric to artificially "flatten" it at some radius $R$. For example, you could take the function $$f(r) = 1 - \frac{r_s}{r}$$ that appears in the metric and replace it by $$\tilde{f}(r) = 1 - \frac{r_s}{r} e^{-r^2/R^2}.$$ For small $r$ this is very close to the Schwarzschild metric, and near $r \approx R$ it goes quickly to 1. You could replace the exponential by any function that goes from 1 to 0 at some given value.

$\endgroup$
2
  • $\begingroup$ Thanks for the suggestion, I have a some Christoffel symbol calculations to do. $\endgroup$ Commented Nov 28, 2021 at 20:43
  • $\begingroup$ Thanks - it worked. I used $e^{-b*r^n}}$, and got something much smoother. $\endgroup$ Commented Nov 29, 2021 at 0:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.